*3 i 9 (1,) i (i,) (1,) 9 (i,) i i 2 1 ( 1, ) (1,) 18 2 i, 2 i i r 3r + 4i 1 i 1 i *4 1 i 9 i 1 1 i i 3 9 +
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1 1 2 IT 1 * π i 1i 2i 3i πi i 2 1 * i π ei *1 *2 x = x 2 + x + 1 = 2 3 1
2 *3 i 9 (1,) i (i,) (1,) 9 (i,) i i 2 1 ( 1, ) (1,) 18 2 i, 2 i i r 3r + 4i 1 i 1 i *4 1 i 9 i 1 1 i i i 3 + 4i 3i i 3 i 9 *5 i 9 i i i 1 2i 2 i 3 *3 * n n 1 *5 (2 + 3i)(4 + 5i) = i + 1i = i r 1 (cos θ 1 + i sin θ 1 ) r 2 (cos θ 2 + i sin θ 2 ) = r 1 r 2 {cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 )} 2
3 1 + 2i 1 i i i 1 *6 1 *7 (1 + i)(a + bi) = a b + (a + b)i ( 1 + i)(a + bi) = a b + (a b)i ( 1 i)(a + bi) = a + b + ( a b)i (1 i)(a + bi) = a + b + ( a + b)i *8 *9 * = (1 + i)(1 i) i 2 = 1 (1 + i)(1 i) = 1 2 i 2 = 1 2 ( 1) = i 1 i i 1 + i 1( 4 ) 1 + i i i 1 + i 1 + i a + bi a, b b a i * , 1, i, i * 12 *6 1 *7 a + bi a 2 + b 2 = 1 4 *8 *9. 2 1(mod 4). 4 1(mod 8) *1 * i = 2 2 = i *12 *13 n n = 1 n = 2, 3, 7, 11, 19, 43, 67, 163 3
4 * 13 2 (a + bi)(a bi) (1 + 2i)(1 2i) = 5 (1 + 4i)(1 4i) = 17 (1 + 6i)(1 6i) = 37 (2 + 3i)(2 3i) = 13 (2 + 5i)(2 5i) = 29 (2 + 7i)(2 7i) = 53 (4 + 5i)(4 5i) = * i * 15 1) 4 3 2) 4 3 2) a + bi a 2 +b 2 * 16 2 * 17 b ) 2 2 2) a 2 + b 2 2 a, b 5 2 = = = = = = = = = = = = = ) 4 1 a * 18 x x log x x x log x 2ax log x x + 2ax log x a a x a * *15 *16 a + bi a bi a bi *17 *18 4
5 3 5 (a + bi)(c + di) = (ac bd) + (ad + bc)i a + bi a 2 + b 2 c + di c 2 + d 2 (ac bd) + (ad + bc)i (a 2 + b 2 )(c 2 + d 2 ) A A A A 6(= ) 28(= ) n M(M + 1) 1(= M) 2 (2 n 1) * 19 GIMPS * 2 a, b < i i i 3 a b a 2 + b i 2+i * 21 a 2 + b n a, b n = a 2 + b 2 a + bi n n 1 + 2i 2 + i * 22 * 23 a + bi a a b *19 * (2 n 1) 1 (2 2n + 1) n 2 Bernoulli *21 9 *22 *23 5
6 a + bi a a 1 a + bi * 24 a * 25 n n n n n 1/2 n 1/1.99 n 1/ j = ( 1) 1/ a x x x x * 28 x e ix = cos x + i sin x π 3 2 * 26 * 27 a + bi e a+bi = e a e bi e a+bi = e a (cos b + i sin b) *24 a 1 a + bi *25 *26 *27 *28 6
7 e α e β = e α+β c + di e e c+di = e c (cos d + i sin d) e a+bi e c+di = e a+c (cos b + i sin b)(cos d + i sin d) = e a+c (cos(b + d) + i sin(b + d)) = e (a+bi)+(c+di) * 29 2π = m = y i < 2π 2 1 log i = log 1 + πi/2( ) 2 i i = e i(πi/2) = e π/2 ( ) e a x = e x log a A A x = e x log A log A a + bi = e x+yi log(a + bi) = x + yi log x a y b OK 2 n n (n 1) (n 2) = 6 (1/2) π * 3 e x+yi = e x (cos y + i sin y) = a 2 + b 2 ( a/ a 2 + b 2 + ib/ a 2 + b 2) = a + bi Γ(n) = (n 1) = x n 1 e x dx e x a 2 + b 2 x = log ( a 2 + b 2) y cos y = a/ a 2 + b 2 sin y = b/ a 2 + b 2 y y y = y + 2mπ m e x+yi = e x (cos y + i sin y) = a + bi x, y ( ) x+yi = log(a+bi) = log a2 + b 2 +i(sin y +2mπ), m log(a + bi) m y m y Γ(n) = (n 1)Γ(n 1) 1 2 n * 31 n = 2.46 Γ(1/2) π Γ(π) Γ(1 + i) = i *29 e *3 e ax dx = 1/a e ax xe ax dx = 1/a 2 *31 n n 7
8 Gamma(n) a aπ/2 a a a sin(n) sin(n) sin(n) = ( 1) x n 2x+1 (2x + 1)! x= = n n3 6 + n5 12 n n n 1 n n x n x n 1 nx n 1 * i d 1 sin(n) dn = cos(n) = ( 1) x n 2x (2x)! x= = 1 n2 2 + n4 24 n n x n 1 nx n 1 n(n 1)x n 2 n(n 1) n 1 n 1 Γ.5.5π/2 sin(n + π/4) n(n 1)(n 2)x n 3 a a x n a = Γ(n + 1) Γ(n + 1 a) xn a sin cos * 33 sin(n) 1 cos(n) sin π/2 9 cos 1 π/2.5 π/4 d.5 sin(n) dn = ( 1) x n 2x+1.5 Γ(2x + 2.5) x= 1 * 34 n * 35 a n *32 *33 *34 *35 n n = 5 8
9 d.5 sin(n) * 36 π 2 2nπ 1 n n n dy/dx n n n n n n n 3 n n 3 π/3 π/2 n π 2π/n (π 2) (π 2) n n 1 9 π/2 2. 1:1 π/ π/ n i 9 (π/2 ) 9 e ix x i e 1 1 (.368 ) i sin(a + bi) cos(ix) sin(ix) *36 π 2π/ 9
10 * 37 cos(ix) = e x + e x 2 sin(ix) = e x e x 2i = cosh x = i sinh x a + bi a + bi 2π a + bi π π 2π/(a + bi) a + bi 1 * 38 ( ) π = 2 sin a + bi A + Bi exp (i(a + Bi)) A e B 4 mod 1 mod 3 1/3 1 mod 1 mod mod 2 2π mod 2π sin(x) cos(x) mod * 39 x 1, x 2, x 3 cx 4 i s s = x x2 2 + x2 3 c2 x 2 4 x 4 i 2 = 1 x 4 s x 4 * 4 mod *37 x ix *38 *39 *4 s s *41 1 1
11 2 ( ) * 41 ( ) i 8i 2 2 * 42 a a n n n 1 1 a L(a) L(a) = a 1 n n 1 1/3 4/3 L(1/3) = (1/3) 1 n = 4/3 n n = log 3 4 = a a c+bi c + bi x 3 3 x 3 x 3 3 Z 3 R 3 C n * 43 n NANIKA n n NANIKA 3 3 n n n Γ x n n a c+bi a c+bi c b x 2i +3x i +5x i x 2 +x 1 +6x = 2 3 a 1x 2i + a 2x 2 + a 3x 1.99i + a 4x *42 *43 11
12 x i = x 1 = x = x i = x i = e i log x = e i e log x = e i e log x = x x i = 1 e log x = 1/e i log x = i x = a + bi log x = log(a + bi) = log( a 2 + b 2 ) + i(sin y + 2mπ) = i log ( a 2 + b 2) = a 2 + b 2 = 1 a, b cos( 1) = a sin( 1) = b x = cos( 1) + i sin( 1) x = e i x 1 = 1 x = 1 = e x i = n log x = i log n e i log n x i = n (x i n)(x i m)(x i l) = n, m, l K mixi PDF D 19 / ISBN C741 ` c Copyright Printed in Japan 12
29
9 .,,, 3 () C k k C k C + C + C + + C 8 + C 9 + C k C + C + C + C 3 + C 4 + C 5 + + 45 + + + 5 + + 9 + 4 + 4 + 5 4 C k k k ( + ) 4 C k k ( k) 3 n( ) n n n ( ) n ( ) n 3 ( ) 3 3 3 n 4 ( ) 4 4 4 ( ) n n
ax 2 + bx + c = n 8 (n ) a n x n + a n 1 x n a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 4
20 20.0 ( ) 8 y = ax 2 + bx + c 443 ax 2 + bx + c = 0 20.1 20.1.1 n 8 (n ) a n x n + a n 1 x n 1 + + a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 444 ( a, b, c, d
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,,,,.,,,. R f : R R R a R, f(a + ) f(a) lim 0 (), df dx (a) f (a), f(x) x a, f (a), f(x) x a ( ). y f(a + ) y f(x) f(a+) f(a) f(a + ) f(a) f(a) x a 0 a a + x 0 a a + x y y f(x) 0 : 0, f(a+) f(a)., f(x)
高等学校学習指導要領解説 数学編
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arctan 1 arctan arctan arctan π = = ( ) π = 4 = π = π = π = =
arctan arctan arctan arctan 2 2000 π = 3 + 8 = 3.25 ( ) 2 8 650 π = 4 = 3.6049 9 550 π = 3 3 30 π = 3.622 264 π = 3.459 3 + 0 7 = 3.4085 < π < 3 + 7 = 3.4286 380 π = 3 + 77 250 = 3.46 5 3.45926 < π < 3.45927
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#A A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/
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![, 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x](/thumbs/93/113428934.jpg ", 1 ( f n (x))dx d dx ( f n (x)) 1 f n (x)dx d dx f n(x) lim f n (x) = [, 1] x f n (x) = n x x 1 f n (x) = x f n (x) = x 1 x n n f n(x) = [, 1] f n (x")
1 1.1 4n 2 x, x 1 2n f n (x) = 4n 2 ( 1 x), 1 x 1 n 2n n, 1 x n n 1 1 f n (x)dx = 1, n = 1, 2,.. 1 lim 1 lim 1 f n (x)dx = 1 lim f n(x) = ( lim f n (x))dx = f n (x)dx 1 ( lim f n (x))dx d dx ( lim f d
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70 : 0 : A B (0 ) (30 ) 50 1 1 4 1.1................................................ 5 1. A............................................... 6 1.3 B............................................... 7 8.1 A...............................................
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No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y
No1 1 (1) 2 f(x) =1+x + x 2 + + x n, g(x) = 1 (n +1)xn + nx n+1 (1 x) 2 x 6= 1 f 0 (x) =g(x) y = f(x)g(x) y 0 = f 0 (x)g(x)+f(x)g 0 (x) 3 (1) y = x2 x +1 x (2) y = 1 g(x) y0 = g0 (x) {g(x)} 2 (2) y = µ
II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2
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II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh
I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google
I4 - : April, 4 Version :. Kwhir, Tomoki TA (Kondo, Hirotk) Google http://www.mth.ngoy-u.c.jp/~kwhir/courses/4s-biseki.html pdf 4 4 4 4 8 e 5 5 9 etc. 5 6 6 6 9 n etc. 6 6 6 3 6 3 7 7 etc 7 4 7 7 8 5 59
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6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P
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I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )
I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17
t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ
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4 4. 4.. 5 5 0 A P P P X X X X +45 45 0 45 60 70 X 60 X 0 P P 4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P 0 0 + 60 = 90, 0 + 60 = 750 0 + 60 ( ) = 0 90 750 0 90 0
> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3
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f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a
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高校生の就職への数学II
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知能科学：ニューラルネットワーク
2 3 4 (Neural Network) (Deep Learning) (Deep Learning) ( x x = ax + b x x x ? x x x w σ b = σ(wx + b) x w b w b .2.8.6 σ(x) = + e x.4.2 -.2 - -5 5 x w x2 w2 σ x3 w3 b = σ(w x + w 2 x 2 + w 3 x 3 + b) x,
1 n A a 11 a 1n A =.. a m1 a mn Ax = λx (1) x n λ (eigenvalue problem) x = 0 ( x 0 ) λ A ( ) λ Ax = λx x Ax = λx y T A = λy T x Ax = λx cx ( 1) 1.1 Th
1 n A a 11 a 1n A = a m1 a mn Ax = λx (1) x n λ (eigenvalue problem) x = ( x ) λ A ( ) λ Ax = λx x Ax = λx y T A = λy T x Ax = λx cx ( 1) 11 Th9-1 Ax = λx λe n A = λ a 11 a 12 a 1n a 21 λ a 22 a n1 a n2
( ) ( ) ( ) i (i = 1, 2,, n) x( ) log(a i x + 1) a i > 0 t i (> 0) T i x i z n z = log(a i x i + 1) i=1 i t i ( ) x i t i (i = 1, 2, n) T n x i T i=1 z = n log(a i x i + 1) i=1 x i t i (i = 1, 2,, n) n
n Y 1 (x),..., Y n (x) 1 W (Y 1 (x),..., Y n (x)) 0 W (Y 1 (x),..., Y n (x)) = Y 1 (x)... Y n (x) Y 1(x)... Y n(x) (x)... Y n (n 1) (x) Y (n 1)
D d dx 1 1.1 n d n y a 0 dx n + a d n 1 y 1 dx n 1 +... + a dy n 1 dx + a ny = f(x)...(1) dk y dx k = y (k) a 0 y (n) + a 1 y (n 1) +... + a n 1 y + a n y = f(x)...(2) (2) (2) f(x) 0 a 0 y (n) + a 1 y
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Chap11.dvi
. () x 3 + dx () (x )(x ) dx + sin x sin x( + cos x) dx () x 3 3 x + + 3 x + 3 x x + x 3 + dx 3 x + dx 6 x x x + dx + 3 log x + 6 log x x + + 3 rctn ( ) dx x + 3 4 ( x 3 ) + C x () t x t tn x dx x. t x
漸化式のすべてのパターンを解説しましたー高校数学の達人・河見賢司のサイト
https://www.hmg-gen.com/tuusin.html https://www.hmg-gen.com/tuusin1.html 1 2 OK 3 4 {a n } (1) a 1 = 1, a n+1 a n = 2 (2) a 1 = 3, a n+1 a n = 2n a n a n+1 a n = ( ) a n+1 a n = ( ) a n+1 a n {a n } 1,
main.dvi
6 FIR FIR FIR FIR 6.1 FIR 6.1.1 H(e jω ) H(e jω )= H(e jω ) e jθ(ω) = H(e jω ) (cos θ(ω)+jsin θ(ω)) (6.1) H(e jω ) θ(ω) θ(ω) = KωT, K > 0 (6.2) 6.1.2 6.1 6.1 FIR 123 6.1 H(e jω 1, ω
di-problem.dvi
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1 filename=mathformula tex 1 ax 2 + bx + c = 0, x = b ± b 2 4ac, (1.1) 2a x 1 + x 2 = b a, x 1x 2 = c a, (1.2) ax 2 + 2b x + c = 0, x = b ± b 2
filename=mathformula58.tex ax + bx + c =, x = b ± b 4ac, (.) a x + x = b a, x x = c a, (.) ax + b x + c =, x = b ± b ac. a (.3). sin(a ± B) = sin A cos B ± cos A sin B, (.) cos(a ± B) = cos A cos B sin
newmain.dvi
数論 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/008142 このサンプルページの内容は, 第 2 版 1 刷発行当時のものです. Daniel DUVERNEY: THÉORIE DES NOMBRES c Dunod, Paris, 1998, This book is published
春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim n an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16,
春期講座 ~ 極限 1 1, 1 2, 1 3, 1 4,, 1 n, n n {a n } n a n α {a n } α {a n } α lim an = α n a n α α {a n } {a n } {a n } 1. a n = 2 n {a n } 2, 4, 8, 16, 32, n a n {a n } {a n } 2. a n = 10n + 1 {a n } lim an
Email: kawahiraamath.titech.ac.jp (A=@) 27 2 25 . 2. 3. 4. 5. 3 4 5 ii 5 50 () : (2) R: (3) Q: (4) Z: (5) N: (6) : () α: (2) β: (3) γ, Γ: (4) δ, : (5) ϵ: (6) ζ: (7) η: (8) θ, Θ: (9) ι: (0) κ: () λ, Λ:
1
1 1 7 1.1.................................. 11 2 13 2.1............................ 13 2.2............................ 17 2.3.................................. 19 3 21 3.1.............................
BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B
2000 8 3.4 p q θ = 80 B E a H F b θ/2 O θ/2 D A B E BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF :
x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin
2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ